You have 1.50 kg of water at$\mathbf{28}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$in an insulated container of negligible mass. You add 0.600 kg of ice that is initially at$\mathbf{-}\mathbf{22}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}$. Assume that no heat exchanges with the surroundings. (a) After thermal equilibrium has been reached, has all of the ice melted? (b) If all of the ice has melted, what is the final temperature of the water in the container? If some ice remains, what is the final temperature of the water in the container, and how much ice remains?

Mass of water: 150 kg

Temperature of water: $28°C$

Mass of ice: 0.600 kg

Temperature of ice: $-22°C$

When two systems are connected to each other via a path that permits heat flow. A point at which the heat flow stops is called as thermal equilibrium

Heat from water when cooled at $0°C$is given by,

$Q=mc∆T$

Here, m is the mass of water, c is the specific heat and $∆T$is the change in temperature.

Substitute values in the above,

$$\begin{gathered} Q=\left(1.5kg\right)\left(4190J/kg.K\right)\left(28K\right) \\ =1.76\times {10}^{5}J \end{gathered}$$

Ice will melt by the hear given,

$Q=mc∆T+m{L}_f$

Here, m is the mass of water, c is the specific heat, is the change in temperature and is the latent heat of ice.

Substitute all the values in the above,

$$\begin{gathered} Q=\left(0.600kg\right)\left[\left(2100J/kg.K\right)\left(22K\right)+3.34\times {10}^{5}J/kg\right] \\ =2.276\times {10}^{5}J \end{gathered}$$

Thus, the heat required is $2.276\times {10}^{5}J$.

For the melting of ice,

$$\begin{gathered} Q_{fromwater}=mc∆T+m_{ice}{L}_f \\ {m}_{ice}=\frac{Q_{fromwater}-mc∆T}{L_f} \end{gathered}$$

Here, m is the mass of water, c is the specific heat, $∆T$is the change in temperature and $L_f$is the latent heat of ice.

Substitute all the values in the above,

$$\begin{gathered} m_{ice}=\frac{1.76\times {10}^{5}J-\left(0.600kg\right)\left(2100J/kg.K\right)\left(22K\right)}{3.34\times {10}^{5}J/kg} \\ {m}_{ice}=0.444kg \end{gathered}$$

The ice remaining is given by,

$$\begin{gathered} m_{remaining}=0.600kg=0.444kg \\ =0.156kg \end{gathered}$$

Thus, there will be still ice remaining in the water so the final temperature is $0°C$.